If we do not know any "log laws," let $u=\log(x^{43})$, $dv=dx$. Then
$du=\dfrac{43x^{42}}{x^{43}}\,dx=\dfrac{43}{x}\,dx$, and we can take $v=x$.
Thus our indefinite integral is
$$x\log(x^{43})-\int \frac{43}{x} x\,dx.$$